Potential of point charge, potential from field and vice versa: Question bank

Before you start, study the anchor picture below. The black curve is a potential landscape $V (x)$ — its height above the axis is the value of $V$ at each position $x$ . Three points are marked: a nearly flat spot at bottom-left (small slope ⇒ almost no field), the hilltop at the peak (zero slope ⇒ $E = 0$ even though $V$ is large), and a point on the steep right-hand slope where the single red arrow shows $E = - d V / d x$ pointing down the slope. Re-read this whenever a trap mentions "height" versus "slope":

Figure — Potential of point charge, potential from field and vice versa

True or false — justify

A charge feels no force at a point ⇒ the potential is zero there.

False.

E

is the local slope of

V

;

V

is the height. At the midpoint of two equal

+ q

charges, symmetry cancels the field (

E = 0

) but both charges raise the height, so

V > 0

— a flat hilltop (like the peak in the anchor figure) still has altitude.

Potential is zero at a point ⇒ the field is zero there.

False. At the midpoint between

+ q

and

- q

the heights cancel (

V = 0

), but both fields point the same way (toward

- q

), so

E \neq = 0

— flat value is not flat slope.

For a point charge,

V

and

E

fall off at the same rate.

False.

V \propto 1/ r

but the field magnitude

E \propto 1/ r^{2}

; the potential falls off slower because it is the accumulated integral of the field, which adds one extra power of

r

Since

V

superposes and

E

superposes, both can be added as plain numbers.

False. Only

V

adds as signed numbers (it is a scalar).

E

superposes as a vector — you must add components; adding magnitudes gives wrong direction and magnitude.

V

is constant along a path, no work is done moving a charge along it.

True. Since

V

is defined as work per unit charge, moving where

V

does not change means work per charge

= Δ V = 0

, so

W = q Δ V = 0

. This is exactly why Equipotential Surfaces cost no work to slide along.

The potential of a point charge can be negative.

True. For

q < 0

V = k q / r < 0

everywhere. Negative

V

means an external agent gains energy bringing a positive test charge

q_{0}

in (the field pulls it).

E = - \nabla V

means the field points toward higher potential.

False. The minus sign flips it:

E

points toward lower potential (downhill, like the red arrow in the anchor figure). Without the minus, a positive charge would spontaneously climb to higher energy, breaking energy conservation.

You can always choose

V = 0

at infinity.

True for localized charges, false for idealized infinite distributions (infinite line/plane). There

\int

from

\infty

diverges, so you must pick a finite reference point instead.

The line integral

- \int_{\infty}^{P} E \cdot d l

gives the same answer along every path.

True in electrostatics. Because

\oint E \cdot d l = 0

(see Conservative Fields and Curl), only the endpoints matter — this is what makes

V (P)

a single-valued function at all.

Doubling the distance from a point charge halves the potential.

True.

V \propto 1/ r

, so

r \to 2 r

gives

V \to V /2

. (Contrast the field magnitude

E \propto 1/ r^{2}

, which drops to a quarter.)

Adding the same constant to

V

everywhere changes the field.

False.

E = - \nabla V

depends only on how

V

varies, not its absolute value;

\nabla (V + C) = \nabla V

, so the field is untouched — this is why the choice of

V = 0

reference is free.

Spot the error

V = k q / r

, so at

r = 0

the potential is infinite — that's the energy stored."

The formula is a mathematical idealization of a point; real charges have finite size, so

r \to 0

never occurs. The divergence signals the model's breakdown, not a physical infinite energy.

E = - \nabla V

, and

V = k q / r

is positive, so

E

must point inward."

Wrong sign logic. Because

E_{r} = - d V / d r

, the minus sign flips the slope:

\frac{d}{d r} (1/ r) = - 1/ r^{2}

, so

E_{r} = - k q \cdot (- 1/ r^{2}) = + k q / r^{2} > 0

for

q > 0

— the field points outward. The sign-flip in

E_{r} = - d V / d r

is exactly what turns a falling potential into an outward field.

"Between two plates,

E = 200

V/m and the plates are 3 m apart, so

V = 200 \times 3

everywhere between."

V

is not a single number in the gap — it varies linearly. Only the difference across 3 m is

600

V; the potential itself depends on which point you're at.

"To get

V

from

E

I differentiate; to get

E

from

V

I integrate."

Backwards.

V

comes from integrating the field (

V = - \int E \cdot d l

);

E

comes from differentiating the potential (

E = - \nabla V

V_{A} - V_{B} = - \int_{A}^{B} E d x

The limits are flipped.

V_{A} - V_{B} = - \int_{B}^{A} E \cdot d l = + \int_{A}^{B} E \cdot d l

. Getting the endpoints wrong flips the sign of the answer.

"The field is strong here, so the potential must be high here."

Field magnitude measures steepness (slope), not height (value). A steep hillside can be at any altitude; a strong field can sit at low, high, or zero potential.

V

is a fixed absolute quantity, so its numerical value at a point is meaningful on its own."

Only differences in

V

are physical. You may add any constant everywhere without changing a single field or force; the raw number depends entirely on where you set

V = 0

Why questions

Why is

V

preferred over

E

for energy calculations?

V

is a scalar, so contributions add as signed numbers with no trigonometry; energy bookkeeping becomes algebra instead of vector addition.

Why does the minus sign appear in

V = - \int E \cdot d l

The external agent must push against the field's force

q_{0} E

on the test charge

q_{0}

; its work is the negative of the field's work, and dividing by

q_{0}

carries that minus into

V

Why is

E

always perpendicular to an equipotential surface?

Along the surface

V

is constant, so its slope in that direction is zero — no field component lies in the surface; all of

E = - \nabla V

must point across it (steepest descent). See Equipotential Surfaces.

Why can we pick a "radial path" when deriving

V

of a point charge?

Because the field is conservative (

\oint E \cdot d l = 0

), the answer is path-independent, so we choose the path where

E ∥ d l

and the integral is easiest.

Why does

E = - \nabla V

recover exactly the

1/ r^{2}

Coulomb field from

V = k q / r

Do the differentiation explicitly:

E_{r} = - \frac{d}{d r} (\frac{k q}{r}) = - k q \cdot (- \frac{1}{r ^{2}}) = \frac{k q}{r ^{2}}

. Differentiating the

1/ r

potential manufactures the extra power of

r

, returning Coulomb's

1/ r^{2}

field magnitude — the bridge is genuinely two-way (see Gradient Operator).

Why is

E = 0

at a genuine maximum or minimum of

V

At a true extremum the slope

d V / d x = 0

in every direction, and since

E = - \nabla V

, the field vanishes there. This is not the same as

V = 0

: the anchor figure's hilltop has zero slope (so

E = 0

) yet a large height (

V \neq = 0

). Zero slope kills the field; zero value does not.

Why is the choice of where

V = 0

arbitrary?

Because physics uses only

E = - \nabla V

and energy differences

q Δ V

; both ignore any constant added to

V

. The reference point is a bookkeeping convenience, not a physical fact.

Edge cases

At the exact midpoint of two equal

+ q

charges (let

d

be the separation between the two charges, so each is a distance

d /2

from the midpoint): what are

V

and

E

E = 0

by symmetry (fields cancel), but

V = 4 k q / d > 0

— each charge at distance

d /2

contributes

k q / (d /2) = 2 k q / d

, and the two add to

4 k q / d

. A nonzero potential with zero field.

At the exact midpoint between

+ q

and

- q

: what are

V

and

E

V = 0

(heights cancel), but

E \neq = 0

— both fields point from

+ q

toward

- q

and add.

What is

V

infinitely far from any finite charge distribution?

V \to 0

, our chosen reference. This is the boundary condition that makes

V = k q / r

come out clean with no added constant.

For a negative charge, which way does

E

point and what is the sign of

V

E

points radially inward (toward the charge) and

V < 0

everywhere; the field still points downhill, from

V = 0

at infinity down into the negative well.

V

is constant throughout a region, what is

E

there?

E = - \nabla V = 0

everywhere in that region — flat ground means no push, e.g. the interior of a charged conductor.

As a test charge

q_{0}

is moved along a closed loop and returns to start, what is the net work per charge?

Zero:

\oint E \cdot d l = 0

for electrostatic fields, so

Δ V = 0

around any closed path — this is what makes

V

well-defined.

Connections

Coulomb's Law — supplies the $1/ r^{2}$ force behind every field-vs-potential trap; the " $r = 0$ is infinite" and " $E \propto 1/ r^{2}$ vs $V \propto 1/ r$ " traps live here.
Electric Field of Point Charge — the vector $E$ we recover by differentiating $V$ ; underpins the "sign of $E_{r} = - d V / d r$ " trap in Spot the error.
Potential Energy of Charge System — $U = q V$ turns the sign-of- $V$ and negative-charge edge cases into energy statements ("negative $V$ ⇒ agent gains energy").
Equipotential Surfaces — the flat-value cases: why "no work along constant $V$ " is True and why $E ⊥$ surface in the Why questions.
Conservative Fields and Curl — the reason $\oint E \cdot d l = 0$ , which justifies the path-independence True/False and the closed-loop edge case.
Gradient Operator — the machinery of $E = - \nabla V$ , invoked by the gauge-freedom ("add a constant") trap and the hilltop $E = 0$ reasoning.